Digital Signal Processing MCQ [Free PDF] – Objective Question Answer for Digital Signal Processing Quiz

The sampling rate conversion technique given in the diagram is decimation by a factor D.

We observe that the sampling rate increase, obtained by the addition of I-1 zero samples between successive values of x(n), results in a signal whose spectrum is an I-fold periodic repetition of the input signal spectrum.

The amplitude of the sampling rate converted signal should be multiplied by a factor C, whose value when equal to I is called the desired normalization factor W.

A sampling rate conversion by the rational factor I/D is accomplished by cascading an interpolator with a decimator.

We emphasize that the importance of performing the interpolation first and decimation second is to preserve the desired spectral characteristics of x(n).

In the diagram given, an interpolator is in a cascade with a decimator which together performs the action of sampling rate conversion by a factor I/D.

We know that the Discrete Fourier transform of a signal x(n) is given as

X(k)=\(\sum_{n=0}^{N-1} x(n)e^{-j2πkn/N}=\sum_{n=0}^{N-1} x(n) W_N^{kn}\)

Thus we get Nth rot of unity WN= e-j2π/N

The formula for calculating N point DFT is given as

X(k)=\(\sum_{n=0}^{N-1} x(n)e^{-j2πkn/N}\)

From the formula given at every step of computing, we are performing N complex multiplications and N-1 complex additions. So, in a total to perform N-point DFT we perform N2 complex multiplications and N(N-1) complex additions.

If XN represents the N point DFT of the sequence xN in the matrix form, then we know that X_N = W_N.x_N
By pre-multiplying both sides by W_N^-1, we get
x_N=W_N-1.X_N

But we know that the inverse DFT of X_N is defined as
x_N=1/N*X_N

Thus by comparing the above two equations we get
W_N-1=1/N W_N*

The first step is to determine the matrix W4. By exploiting the periodicity property of W4 and the symmetry property
\(W_{N}^{k+N/2}=-W_{N^k}\)

The matrix W4 may be expressed as

W4=\(\begin{bmatrix}W_4^0&W_4^0&W_4^0&W_4^1\\W_4^0&W_4^0&W_4^2&W_4^3\\W_4^0&W_4^2&W_4^0&W_4^3\\W_4^4&W_4^6&W_4^6&W_4^9\end{bmatrix}=\begin{bmatrix}W_4^0&W_4^0&W_4^0&W_4^1\\W_4^0&W_4^0&W_4^2&W_4^3\\W_4^0&W_4^2&W_4^0&W_4^3\\W_4^0&W_4^2&W_4^2&W_4^1\end{bmatrix}\)
=\(\begin{bmatrix}1&1&1&1\\1&-j&-1&j\\1&-1&1&-1\\1&j&-1&-j\end{bmatrix}\)

Then X4=W4.x4=\(\begin{bmatrix}6\\-2+2j\\-2\\-2-2j\end{bmatrix}\)

If X(k) is the N point DFT of a sequence the Fourier series coefficients are given by the expression

ck=\(\frac{1}{N} \sum_{n=0}^{N-1} x(n)e^{-j2πkn/N} = \frac{1}{N}X(k)=> X(k)=Nc_k\)

Given x(n)={0,1,2,3}

We know that the 4-point DFT of the above given sequence is given by the expression

X(k)=\(\sum_{n=0}^{N-1} x(n)e^{-j2πkn/N} \)

In this case N=4

=>X(0)=6, X(1)=-2+2j, X(2)=-2, X(3)=-2-2j.

We know that according to the periodicity and symmetry property,
100/4=200/x=>x=8.

Answer: A

As in the analog domain, frequency transformations can be performed on a digital low pass filter to convert it to either a bandpass, band stop, or high pass filter. The transformation involves the replacing of the variable z-1 with a rational function g(z-1).

Answer: C

The map z-1 → g(z-1) must map inside the unit circle in the z-plane into itself to apply digital frequency transformation.

Answer: B

For the map z-1 → g(z-1) to be a valid digital frequency transformation, then the unit circle also must be mapped inside the unit circle.

Answer: D

We know that the unit circle must be mapped inside the unit circle.
Thus it implies that for r=1, e-jω = g(e-jω)=|g(ω)|.ej arg [ g(ω) ]
It is clear that we must have |g(ω)|=1 for all ω. That is, the mapping is all-pass.

Answer: A

The value of |ak| < 1 to ensure that a stable filter is transformed into another stable filter to satisfy condition 1.

Answer: C

We know that the impulse invariance method and mapping of derivatives are inappropriate to use in the designing of high pass and bandpass filters.

Answer: A

We know that the impulse invariance method and mapping of derivatives are inappropriate to use in the designing of high pass and bandpass filters due to the aliasing problems.

Answer: B

Since there is a problem with aliasing in designing high pass and many bandpass filters using impulse invariance and mapping of derivatives, we cannot employ the analog frequency transformation followed by conversion of the result into the digital domain by use of these two mappings.

Answer: A

It is better to perform the mapping from an analog low pass filter into a digital low pass filter by either of these mappings and then perform the frequency transformation in the digital domain because, by this kind of frequency transformation, the problem of aliasing is avoided.

Answer: C

In the case of bilinear transformation, where aliasing is not a problem, it does not matter whether the frequency transformation is performed in the analog domain or in the frequency domain.